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Balancing simple redox reactions can be a straightforward matter of going back and forth between products and reactants. For example, in the redox reaction of Na and Cl₂:

Na + Cl₂ → NaCl

it should be immediately clear that the Cl atoms are not balanced. We can fix this by putting the coefficient 2 in front of the product:

Na + Cl₂ → 2NaCl

However, now the sodium is unbalanced. This can be fixed by including the coefficient 2 in front of the Na reactant:

2Na + Cl₂ → 2NaCl

This reaction is now balanced. That was fairly straightforward; we say that we are able to balance the reaction by inspection. Many simple redox reactions can be balanced by inspection.

The first thing you should do when encountering an unbalanced redox reaction is to try to balance it by inspection.

Some redox reactions are not easily balanced by inspection. Consider this redox reaction:

Al + Ag⁺ → Al³⁺ + Ag

At first glance, this equation seems balanced: there is one Ag atom on both sides and one Al atom on both sides. However, if you look at the total charge on each side, there is a charge imbalance: the reactant side has a total charge of 1+, while the product side has a total charge of 3+. Something is amiss with this chemical equation; despite the equal number of atoms on each side, it is not balanced.

A fundamental point about redox reactions that has not arisen previously is that the total number of electrons being lost must equal the total number of electrons being gained for a redox reaction to be balanced. This is not the case for the aluminum and silver reaction: the Al atom loses three electrons to become the Al³⁺ ion, while the Ag⁺ ion gains only one electron to become elemental silver.

To balance this, we will write each oxidation and reduction reaction separately, listing the number of electrons explicitly in each. Individually, the oxidation and reduction reactions are called half reactions. We will then take multiples of each reaction until the number of electrons on each side cancels completely and combine the half reactions into an overall reaction, which should then be balanced. This method of balancing redox reactions is called the half reaction method. (There are other ways of balancing redox reactions, but this is the only one that will be used in this text. The reason for this will be seen in the section “Applications of Redox Reactions: Voltaic Cells”.)

The oxidation half reaction involves aluminum, which is being oxidized:

Al → Al³⁺

This half reaction is not completely balanced because the overall charges on each side are not equal. When an Al atom is oxidized to Al³⁺, it loses three electrons. We can write these electrons explicitly as products:

Al → Al³⁺ + 3e⁻

Now this half reaction is balanced — in terms of both atoms and charges.

The reduction half reaction involves silver:

Ag⁺ → Ag

The overall charge is not balanced on both sides. But we can fix this by adding one electron to the reactant side because the Ag⁺ ion must accept one electron to become the neutral Ag atom:

Ag⁺ + e⁻ → Ag

This half reaction is now also balanced.

When combining the two half reactions into a balanced chemical equation, the key is that the total number of electrons must cancel, so the number of electrons lost by atoms are equal to the number of electrons gained by other atoms. This may require we multiply one or both half reaction(s) by an integer to make the number of electrons on each side equal. With three electrons as products and one as reactant, the least common multiple of these two numbers is three: we can use a single aluminum reaction but must take three times the silver reaction:

[latex]\begin{align*} \ce{Al}&\rightarrow \ce{Al^{3+}}+3e^- \\ 3&\times [\ce{Ag+}+e^-\rightarrow \ce{Ag}] \end{align*}[/latex]

The 3 on the second reaction is distributed to all species in the reaction:

[latex]\begin{align*} \ce{Al}&\rightarrow \ce{Al^{3+}}+3e^- \\ \ce{3Ag+}&+3e^- \rightarrow \ce{3Ag} \end{align*}[/latex]

Now the two half reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. The same species on opposite sides of the arrow can be cancelled:

[latex]\ce{Al}+\ce{3Ag+}+3e^-\rightarrow \ce{Al^{3+}}+\ce{3Ag}+3e^-[/latex]

The net balanced redox reaction is as follows:

[latex]\ce{Al}+\ce{3Ag+}\rightarrow \ce{Al^{3+}}+\ce{3Ag}[/latex]

There is still only one Al atom on each side of the chemical equation, but there are now three Ag atoms, and the total charge on each side of the equation is the same (3+ for both sides). This redox reaction is balanced. It took more effort to use the half reaction method than by inspection, but the correct balanced redox reaction was obtained.

Many redox reactions occur in aqueous solution — in water. Because of this, in many cases H₂O or a fragment of an H₂O molecule (H⁺ or OH⁻, in particular) can participate in the redox reaction. As such, we need to learn how to incorporate the solvent into a balanced redox equation.

Consider the following oxidation half reaction in aqueous solution, which has one Cr atom on each side:

Cr³⁺ → CrO₄⁻

Here, the Cr atom is going from the +3 to the +7 oxidation state. To do this, the Cr atom must lose four electrons. Let us start by listing the four electrons as products:

Cr³⁺ → CrO₄⁻ + 4e⁻

But where do the O atoms come from? They come from water molecules or a common fragment of a water molecule that contains an O atom: the OH⁻ ion. When we balance this half reaction, we should feel free to include either of these species in the reaction to balance the elements. Let us use H₂O to balance the O atoms; we need to include four water molecules to balance the four O atoms in the products:

4H₂O + Cr³⁺ → CrO₄⁻ + 4e⁻

This balances the O atoms, but now introduces hydrogen to the reaction. We can balance the H atoms by adding an H⁺ ion, which is another fragment of the water molecule. We need to add eight H⁺ ions to the product side:

4H₂O + Cr³⁺ → CrO₄⁻ + 4e⁻ + 8H⁺

The Cr atoms are balanced, the O atoms are balanced, and the H atoms are balanced; if we check the total charge on both sides of the chemical equation, they are the same (3+, in this case). This half reaction is now balanced, using water molecules and parts of water molecules as reactants and products.

Reduction reactions can be balanced in a similar fashion. When oxidation and reduction half reactions are individually balanced, they can be combined in the same fashion as before: by taking multiples of each half reaction as necessary to cancel all electrons. Other species, such as H⁺, OH⁻, and H₂O, may also have to be cancelled in the final balanced reaction.

Unless otherwise noted, it does not matter if you add H₂O or OH⁻ as a source of O atoms, although a reaction may specify acidic solution or basic solution as a hint of what species to use or what species to avoid. OH⁻ ions are not very common in acidic solutions, so they should be avoided in those circumstances.